题目 96 Unique Binary Search Trees I
Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n?
Example:
Input: 3
Output: 5
Explanation:
Given n = 3, there are a total of 5 unique BST's:
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
思路
选择一个数字作为根节点，则其左右两侧子树中节点的个数也确定了。左侧子树 种类数 * 右侧子树种类数 = 该根节点树的个数。为根节点遍历所有数字，得到总树个数。
def numTrees(self, n: int) -> int:
if n==0:
return 0
nums = collections.defaultdict(lambda:0)
nums[0] = 1
nums[1] = 1
for i in range(2,n+1):
if i % 2 == 1:
nums[i] = nums[(i-1)//2]**2
for j in range((i)//2):
nums[i] += 2*nums[j]*nums[i-1-j]
return nums[n]题目 95 Unique Binary Search Trees II
Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.
Example:
Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
思路
用递归，取一个数字作为根节点，所有小于这个数的数组成其左侧的树，大于这个数的数组成其右侧的树。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def generateTree(self,nums):
if len(nums)==0:
return [None]
ans = []
for i in range(len(nums)):
l = self.generateTree(nums[:i])
r = self.generateTree(nums[i+1:])
for li in l:
for ri in r:
root = TreeNode(nums[i])
root.left = li
root.right = ri
ans.append(root)
return ans
def generateTrees(self, n: int) -> List[TreeNode]:
if n==0:
return None
nums = []
for i in range(1,n+1):
nums.append(i)
return self.generateTree(nums)
